If 2vec{a} + 3vec{b} + vec{c} = vec{0},then v | vedclass

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(B) Given,$2\vec{a} + 3\vec{b} + \vec{c} = \vec{0}$.From the given equation,we have $\vec{c} = -(2\vec{a} + 3\vec{b})$.Now,consider the expression $\v

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$3(\vec{b} 	imes \vec{c})$

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(B) Given,$2\vec{a} + 3\vec{b} + \vec{c} = \vec{0}$.From the given equation,we have $\vec{c} = -(2\vec{a} + 3\vec{b})$.Now,consider the expression $\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}$.Substitute $\vec{c} = -(2\vec{a} + 3\vec{b})$ into the expression:$= \vec{a} 	imes \vec{b} + \vec{b} 	imes (-(2\vec{a} + 3\vec{b})) + (-(2\vec{a} + 3\vec{b})) 	imes \vec{a}$Using the distributive property of the cross product:$= \vec{a} 	imes \vec{b} - 2(\vec{b} 	imes \vec{a}) - 3(\vec{b} 	imes \vec{b}) - 2(\vec{a} 	imes \vec{a}) - 3(\vec{b} 	imes \vec{a})$Since $\vec{x} 	imes \vec{x} = \vec{0}$ and $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$:$= \vec{a} 	imes \vec{b} - 2(-(\vec{a} 	imes \vec{b})) - 3(\vec{0}) - 2(\vec{0}) - 3(-(\vec{a} 	imes \vec{b}))$$= \vec{a} 	imes \vec{b} + 2(\vec{a} 	imes \vec{b}) + 3(\vec{a} 	imes \vec{b})$$= 6(\vec{a} 	imes \vec{b})$.Alternatively,from $2\vec{a} + 3\vec{b} + \vec{c} = \vec{0}$,taking cross product with $\vec{b}$:$2(\vec{a} 	imes \vec{b}) + 3(\vec{b} 	imes \vec{b}) + (\vec{c} 	imes \vec{b}) = \vec{0} \Rightarrow 2(\vec{a} 	imes \vec{b}) = \vec{b} 	imes \vec{c} \Rightarrow \vec{a} 	imes \vec{b} = \frac{1}{2}(\vec{b} 	imes \vec{c})$.Substituting this into $6(\vec{a} 	imes \vec{b})$ gives $6 	imes \frac{1}{2}(\vec{b} 	imes \vec{c}) = 3(\vec{b} 	imes \vec{c})$.

If $2\vec{a} + 3\vec{b} + \vec{c} = \vec{0}$,then $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ is equal to

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